sqrt{12-sqrt{140}} = dots | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To simplify the expression $\sqrt{12-\sqrt{140}}$,we first rewrite $\sqrt{140}$ as $2\sqrt{35}$.Thus,the expression becomes $\sqrt{12-2\sqrt{35}}$

Vedclass · Accepted Answer

$\sqrt{7}-\sqrt{5}$

Vedclass · Answer

(C) To simplify the expression $\sqrt{12-\sqrt{140}}$,we first rewrite $\sqrt{140}$ as $2\sqrt{35}$.Thus,the expression becomes $\sqrt{12-2\sqrt{35}}$.We look for two numbers whose sum is $12$ and whose product is $35$. These numbers are $7$ and $5$.We can write $12$ as $(7+5)$ and $35$ as $(7 	imes 5)$.So,$\sqrt{12-2\sqrt{35}} = \sqrt{(7+5)-2\sqrt{7 	imes 5}}$.Using the identity $(a-b)^2 = a^2+b^2-2ab$,we have $\sqrt{(\sqrt{7}-\sqrt{5})^2}$.Therefore,the result is $\sqrt{7}-\sqrt{5}$.

$\sqrt{12-\sqrt{140}} = \dots$

Similar Questions

Euclid's division lemma states that for two positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that $a = bq + r,$ where $r$ must satisfy

If $n$ is an odd integer,then show that $n^{2}-1$ is divisible by $8$.

The conjugate surd of $3+\sqrt{2}$ is $\ldots \ldots \ldots \ldots .$

Prove that the following number is irrational: $\sqrt{5}-\sqrt{3}$.

The decimal form of $\frac{42}{35}$ is $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module